In an algebra test the following problem was presented:
Any commutative local ring $(R,\mathfrak m)$ with $\mathfrak m$ principal so that $⋂_{i≥0}\mathfrak m^i =0$ is Noetherian and each nonzero ideal of $R$ is a power of $\mathfrak m$.
It could be proven (assuming $R$ to be Noetherian) that each nonzero ideal, being finitely generated, is a subset of a power of $\mathfrak m$. Could anybody be so kind as to give some suggestions. Greateful!
Best Answer
Let $I$ be an ideal of $R$, $I\ne 0,R$. Then $I\subseteq\mathfrak m$, but $I$ can't be contained in all the powers of $\mathfrak m$, otherwise $I=0$, contradiction. So there is a greatest $i\ge 1$ such that $I\subseteq \mathfrak m^i$, and $I\not\subseteq\mathfrak m^{i+1}$. Let's show that $I=\mathfrak m^i$. Since $I\not\subseteq\mathfrak m^{i+1}$ there is $a\in I\setminus\mathfrak m^{i+1}$. If $\mathfrak m=(\pi)$, then $a=\pi^ia'$. Since $a\notin\mathfrak m^{i+1}$ we get $a'\notin\mathfrak m$, so $a'$ is invertible. From $a=\pi^ia'$ we get now that $\pi^i\in I$ and therefore $I=\mathfrak m^i$.